@TheChaz It is that time of year when people are getting into courses which deal with this kind of problem, and are coming at it fresh, and don't realise that it was (inevitably) asked last year, and will inevitably be asked again in some equivalent form.
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Mark BennetOct 30 '11 at 21:03

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@Mark: In the preview of your reply, I saw "It is that time of year..." and thought you were going to say something about Xmas... Alas! :)
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The Chaz 2.0Oct 30 '11 at 23:04

2 Answers
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Since $R$ is an integral domain, you only need to show every nonzero element is invertible. Let $x\in R$ be nonzero. Then $x$ induces a map $\phi_x\colon R\to R : r\mapsto xr$, which is injective since $R$ is an integral domain. Thus $\ker\phi_x=\{0\}$, so $\dim\ker\phi_x=0$. Note that $\phi_x$ is a homomorphism, since for any $c\in K$, $r,s\in R$,
$$
\phi_x(cr+s)=x(cr+s)=x(cr)+xs=c(xr)+xs=c\phi_x(r)+\phi_x(s).
$$
By Theorem 5.3 of Lang, $\dim\operatorname{Im}(\phi_x)=\dim R$. By Corollary 5.4, $\operatorname{Im}(\phi_x)=R$, so $\phi_x$ is surjective. So there exists some $y\in R$ such that $\phi_x(y)=xy=1$, and $x$ is invertible, and $R$ is a field.